I have two questions after reading the Hahn-Banach theorem from Conway's book ( I have googled to know the answer but I have not found any result yet. Also I am not sure that whether my questions have been asked here somewhere on this forum - so please feel free to delete them if they are not appropriate )

Here are my questions:

We know that if $M$ is a linear subspace of $X$ and $f :M\to\mathbb{F}$ and $f$ is linear,bounded by a seminorm $p$ then $f$ can be extended onto $X$ by some functional $F$. Can $F$ be unique ? Under what condition $F$ will be an unique extension? It would be appreciate if you could give me one example that $F$ could not be unique.

If the above $\mathbb{F}$ is replaced a Banach space $Y$, i.e, let $M$ be a closed subspace of a Banach space $X$, and $f :M\to Y$ be a bounded linear operator, can we extend $f$ by a bounded operator $F :X\to Y$ ? if not, what condition should be put on $Y$ to have a such extension?

I think that the correct spelling is "bounded".... "bonded" reminds me of bondage.
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André HenriquesJun 25 '11 at 22:44

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Steven, for your 2nd question one needs to put conditions on $M$, $X$ or $Y$. The classic example is to take $M=c_0$ sitting inside $X=\ell^\infty$, take $Y=c_0$, and let $f$ be the identity mapping; then there is no bounded lin. extension of $f$ to a map $X\to Y$, because that would give a bounded lin. projection of $\ell^\infty$ onto $c_0$, which is impossible by a theorem of Phillips. I recommend that you look into a text on Banach space theory: I haven't read Albiac and Kalton's book but my guess is that they might discuss issues like this, and the text itself is at an introductory level.
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Yemon ChoiJun 25 '11 at 23:19

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In question 1, if $X=\mathbb R^2$ with norm $\|(x_1,x_2)\| = \max(x_1,x_2)$, then extension is not unique.
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Gerald EdgarJun 25 '11 at 23:29

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For question 2 it is enough for $M$ to be a complemented subspace, i.e., that it has a closed complement in $X$, since this is equivalent to having a continuous projection from $X$ onto $M$. If the extension property has to hold for every $Y$ and every $f$, then this condition is also necessary, as can be seen by taking $f$ to be the identity map on $M$.
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godelianJun 26 '11 at 0:14

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Steven: I forget the exact wording of the "Phillips lemma", but it is almost surely mentioned in the Kalton-Albiac book (or indeed others)
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Yemon ChoiJun 26 '11 at 1:15

3 Answers
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$Y$ is called an injective Banach space if the extension exists for all $X$, $M$, and $f$. An example is $Y = l^\infty$. (Should be in Banach space text books. Here's a paper: http://www.jstor.org/pss/1998210 )

Continuous extensions of (continuous) functionals from $M$ are unique if and only if $M$ is a dense subspace of $X$. Otherwise its closure is a proper closed subspace and therefore there exists a nonzero bounded functional $\phi$ vanishing on the closure, which implies that $F+\phi$ is bounded and also extends $f$.

For the second question, it is easier to put conditions on $M$ so that for every $Y$, every map from $M$ to $Y$ can be extended. As mentioned in the comments, a necessary and sufficient condition is that $M$ is a complemented subspace of $X$.

Sorry, I only realized after this answer was posted that the question is about extensions bounded by seminorms, not just continuous extensions (the most commonly used corollary of Hahn-Banach theorem). This should have been a comment.
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godelianJun 26 '11 at 15:28

When $X^*$ is strictly convex and $M$ is a subspace of $X$, every norm one $f\in M^*$ admits a unique norm one extension $F\in X^*$, because given two of such extensions $F_1$ and $F_2$, $(F_1+F_2)/2$ is norm one.